Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{4r^2 + 28r}{r^2 - 10r} \times \dfrac{4r^2 - 44r + 120}{r^2 + r - 42} $
Solution: First factor out any common factors. $x = \dfrac{4r(r + 7)}{r(r - 10)} \times \dfrac{4(r^2 - 11r + 30)}{r^2 + r - 42} $ Then factor the quadratic expressions. $x = \dfrac {4r(r + 7)} {r(r - 10)} \times \dfrac {4(r - 6)(r - 5)} {(r - 6)(r + 7)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {4r(r + 7) \times 4(r - 6)(r - 5) } {r(r - 10) \times (r - 6)(r + 7) } $ $x = \dfrac {16r(r - 6)(r - 5)(r + 7)} {r(r - 6)(r + 7)(r - 10)} $ Notice that $(r - 6)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {16r\cancel{(r - 6)}(r - 5)(r + 7)} {r\cancel{(r - 6)}(r + 7)(r - 10)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $x = \dfrac {16r\cancel{(r - 6)}(r - 5)\cancel{(r + 7)}} {r\cancel{(r - 6)}\cancel{(r + 7)}(r - 10)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $x = \dfrac {16r(r - 5)} {r(r - 10)} $ $ x = \dfrac{16(r - 5)}{r - 10}; r \neq 6; r \neq -7 $